Solve the differential equation $\;\;$ $y + 2 \; y \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; x \; e^{x/y}$
The given differential equation is $\;\;$ $y + 2 \; y \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; x \; e^{x/y}$
i.e. $1 + 2 \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; \dfrac{x}{y} \; e^{x/y}$
i.e. $2 \; e^{x/y} \; \dfrac{dx}{dy} = 2 \; \dfrac{x}{y} \; e^{x/y} - 1$ $\;\;\; \cdots \; (1)$
Let $\dfrac{x}{y} = v$ $\;\;\; \cdots \; (2a)$
i.e. $x = v \; y$ $\;\;\; \cdots \; (2b)$
Differentiating equation $(2b)$ w.r.t $y$ gives
$\dfrac{dx}{dy} = v + y \; \dfrac{dv}{dy}$ $\;\;\; \cdots \; (2c)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ becomes
$2 \; e^v \left(v + y \; \dfrac{dv}{dy}\right) = 2 \; v \; e^v - 1$
i.e. $v + y \; \dfrac{dv}{dy} = \dfrac{2 \; v \; e^v - 1}{2 \; e^v}$
i.e. $v + y \; \dfrac{dv}{dy} = v - \dfrac{1}{2 \; e^v}$
i.e. $y \; \dfrac{dv}{dy} = \dfrac{-1}{2 \; e^v}$
i.e. $2 \; e^v \; dv = - \dfrac{dy}{y}$ $\;\;\; \cdots \; (3)$
Integrating equation $(3)$ gives
$2 \displaystyle \int e^v \; dv = - \int \dfrac{dy}{y}$
i.e. $2 \; e^v = - \log \left|y\right| + \log \left|c\right|$ $\;\;\; \cdots \; (4)$
Substituting the value of $v$ from equation $(2a)$ in equation $(4)$ gives
$2 \; e^{x/y} = \log \left|\dfrac{c}{y}\right|$ $\;\;\; \cdots \; (5)$
Equation $(5)$ is the general solution of the given differential equation.